Transcendental Extensions
Both for their own sake and for applications to the case of finite extensions of the rational numbers, one is led to deal with ground fields which are function fields, i.e. finitely generated over some field k, possibly by elements which are not algebraic
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VIII
Transcendental Extensions
Both for their own sake and for applications to the case of finite extensions of the rational numbers, one is led to deal with ground fields which are function fields, i.e. finitely generated over some field k, possibly by elements which are not algebraic. This chapter gives some basic properties of such fields.
§1.
TRANSCENDENCE BASES
Let K be an extension field of a field k. Let S be a subset of K . We recall that S (or the elements of S) is said to be algebraically independent over k, if whenever we have a relation
with coefficients Q(v) E k, almo st all Q(v) = 0, then we must necessarily have all = 0. We can introduce an ordering among algebraically independent subsets of K, by ascending inclusion. These subsets are obviously inductively ordered, and thus there exist maximal elements. If S is a subset of K which is algebraically independent over k, and if the cardinality of S is greatest among all such subsets, then we call this cardinality the transcendence degree or dimension of Kover k. Actually, we shall need to distinguish only between finite transcendence degree or infinite transcendence degree. We observe that
Q(v)
355 S. Lang, Algebra © Springer Science+Business Media LLC 2002
356
TRANSCENDENTAL EXTENSIONS
VIII , §1
the notion of transcendence degree bears to the notion of algebraic independence the same relation as the notion of dimension bears to the notion of linear independence. We frequently deal with families of elements of K , say a family {Xd iel> and say that such a family is algebraically independent over k if its elements are distinct (in other words, Xi =F Xj if i =F j) and if the set consisting of the elements in this family is algebraically independent over k. A subset S of K which is algebraically independent over k and is maximal with respect to the inclusion ordering will be called a transcendence base of Kover k. From the maximality, it is clear that if S is a transcendence base of Kover k, then K is algebraic over k(S).
Theorem 1.1. Let K be an extension of a field k. Any two transcendence bases of Kover k have the same cardinality. If r is a subset of K such that K is algebraic over k(f), and S is a subset of r which is algebraically independent over k, then there exists a transcendence base CB of Kover k such that S C CB Cr.
Proof. We shall prove that if there exists one finite transcendence base, say {x I> • • • ,xm } , m ~ 1, m minimal , then any other transcendence base must also have m elements. For this it will suffice to prove: If WI' ••. , W n are elements of K which are algebraically independent over k then n ~ m (for we can then use symmetry) . By assumption, there exists a non-zero irreducible polynomial fl in m + 1 variables with coefficients in k such that fl(wl'
XI ' • . •
, xm )
= O.
After renumbering x I ' • .• , Xm we may write fl = ~ gi WI ' X2' . . . , x m ) x1 with some gN "* 0 with some N ~ 1. No irreducible factor of gN vanishes on (wI' X2' . .. ,xn ) , otherwise WI would be a root of two distinct irreducible pol
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